Question: $f(n) = 6n^{2}+4n-6$ $g(t) = 3t^{2}+3(f(t))$ $h(t) = t^{2}+2t+2+4(f(t))$ $ f(g(-1)) = {?} $
Solution: First, let's solve for the value of the inner function, $g(-1)$ . Then we'll know what to plug into the outer function. $g(-1) = 3(-1)^{2}+3(f(-1))$ To solve for the value of $g$ , we need to solve for the value of $f(-1)$ $f(-1) = 6(-1)^{2}+(4)(-1)-6$ $f(-1) = -4$ That means $g(-1) = 3(-1)^{2}+(3)(-4)$ $g(-1) = -9$ Now we know that $g(-1) = -9$ . Let's solve for $f(g(-1))$ , which is $f(-9)$ $f(-9) = 6(-9)^{2}+(4)(-9)-6$ $f(-9) = 444$